Question

1 Predict the major product expected with correct stereochemistry (if any) in each of the following reactions (14 points). Ph Br2, NaOH 1) LiCu(CH3)2 2) CH3I

          1 Predict the major product expected with correct stereochemistry (if any) in each of the following reactions (14 points).
Ph
Br2, NaOH
1) LiCu(CH3)2
2) CH3I

1 Predict the major product expected with correct stereochemistry (if any) in each of the following reactions (14 points).
Ph
Br2, NaOH
1) LiCu(CH3)2
2) CH3I

Added by Alejandra A.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Supreeta N

04/28/2023

Step 1

First, Br2 and NaOH will react in a halohydrin formation reaction. This will result in the formation of a bromohydrin with the bromine and hydroxyl group on the same carbon atom. Show more…

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Predict the major product expected with correct stereochemistry (if any) in each of the following reactions (14 points).

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Transcript

00:01 Hello students, in this question we have been given two chemical reactions and we have to form major products.

00:07 First reaction is phenyl this is phenyl methyl ketone it is reacting with br2 in presence of naoh.

00:19 So this is an oxidizing agent it will convert it into carboxylic acid.

00:24 Let's write its mechanism.

00:26 We know that protons at alpha carbon to carbonyl group are acidic hence this naoh will abstract these protons and these will be replaced by bromine hence the product that will be formed after three steps will look like this.

00:50 All these three protons will be removed by bromine atoms after addition of three naoh.

00:58 Now since base is still present and this carbon atom contains three electronegative elements it has become a very good leaving group.

01:08 So in the next step we will show the nucleophilic attack of this hydroxyl ion on this carbonyl carbon as a result of which this carbonyl bond will be cleaved towards oxygen atom.

01:21 So this will be formed.

01:27 Now we will show that these electrons will form the pi bond again and the cbr3 will be removed as a leaving group.

01:39 Hence benzoic acid will be formed.

01:45 So this will be the product that will be formed in this reaction.

01:50 Now in second reaction we have given an alpha beta unsaturated ketone.

02:01 So this is the ketone whose alpha and beta positions are unsaturated and we have to react it in the first step with gilman's reagent is cuch3o.

02:19 In the second step we have to treat it with methyl iodide.

02:25 So gilman's reagent is an alkylating agent which gives ch3 negative.

02:31 Now since this gilman's reagent is a soft reagent because of copper metal if we compare it with grignard's reagent it contains magnesium metal.

02:46 So grignard's reagent is hard and this is a soft reagent.

02:52 So it will attack on soft electrophilic site.

02:55 Now if we talk about alpha beta unsaturated ketones this is known as a hard electrophilic site and this fourth carbon is known as a soft electrophilic site.

03:07 So the methyl group that will be liberated will attack on the soft side and as a result this bond will be shifted here...

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